Thermodynamics - Calculate the initial temperature of the iron.

  • Thread starter SAFiiNA
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  • #1
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Homework Statement



A 100g of hot iron rivets is heated to some temperature and then quickly transferred to 120g of water at 40°C in a copper calorimeter of mass 50g and specific heat capacity of 400J/(kg.°C).It is found that the final temperature of the mixture is 100°C and the water has lost20% of its mass to steam. If the iron has a heat capacity of 460 J/(kg.°C), calculate the initial temperature of the iron. (The latent heat of vapour is 2.26 m/kg)


Homework Equations



E= M x C x ∆T and E = M x L Are these the equations I am supposed to use?

The Attempt at a Solution



I don’t want an answer to this. There is lot of data in this question which is confusing me. Can anyone please break the question down for me into simpler form? Thank you

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Delphi51
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Heat lost by iron = heat gained by warming the water + heat used to make steam
 
  • #3
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Heat lost by iron = heat gained by warming the water + heat used to make steam

Thank you Delphi51. :) I really appreciate your help. In the question they have given mass and heat capacity of copper calorimeter. I don't know what I need that information for and what I am supposed to work out first. :confused: Confused.com
I really want to work thisout myself but I am not gettin anywhere with this.
 
  • #4
Delphi51
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I missed that! Good catch.

Heat lost by iron = heat gained by water + heat gained by calorimeter + heat used to make steam
 
  • #5
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I missed that! Good catch.

Heat lost by iron = heat gained by water + heat gained by calorimeter + heat used to make steam

Thank you so much. I know how to work it out now. This is going to be the last question (hopefully)
In the question it says iron quickly tranferred to water at 40°C in a calorimeter of mass 50g....
I will be using E=MxCxΔT for heat gained by water and calorimeter and add them together. Do I use that 40°C in this equation for both of them?
Like 120x4.2x(100-40) for water and 50x400x(100-40) for the container?
 
  • #6
Delphi51
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Yes, the 100-40 looks good for both. 4.2 for the heat capacity does not look right. I seem to remember 4200. Better check it.
 

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